Simple but interesting mathematics

On cube roots of the cubes of two-digit integers.

This problem appeared on an Australian television
program "The Panel" in 2001. A guest was able to determine, in his
head, the cube root of the cube of any two-digit integer. For example,
a member of the panel would choose a two digit integer (59 say), determine
its cube using a calculator (59x59x59 = 205379) and would then give this
number to the guest, who would instantly respond with its cube root, 59.
How did the guest do this?

Actually, its very simple (of course it had to
be, or else the guest wouldn't have been able to do it!). It relies
on the fact that there is a one-to-one mapping between integers 0 to 9
and the last digit of their cubes. i.e. 0^3 = 0
0->0 1^3 = 1
1->1 2^3 = 8
2->8 3^3 = 27
3->7 4^3 = 64
4->4 5^3 = 125
5->5 6^3 = 216
6->6 7^3 = 343
7->3 8^3 = 512
8->2 9^3 = 729
9->9The mapping turns out to be very easy to remember
also, when one notices that the last digit of the cube is the same as the
original for 0,1,4,5,6, and 9, and is the 10's complement for 2,3,7 and
8. That is, given the last digit of the cube, one can instantly determine
the last digit of the original two-digit integer. The first digit
is of course determined by the range in which the cube falls.

e.g. 79507 lies between 64000 and 125000, and
therefore the first digit of the two-digit integer is 4. The last digit
of the cube is 7, so the last digit of the two-digit integer is 3.
That is 79507^{1/3} = 43.

I wonder what this fellow would have thought the
cube root of 79508 was. I bet he would have said 42, which is of
course nonsense.

As an aside, similar one-to-one mappings exist
for all odd-numbered powers. The above mapping holds for powers of
3, 7, 11, ... For powers of 5, 9, 13, ... the mapping is even simpler,
with the last digit of the result being the same as the original integer.
i.e.0->01->12->2...9->9So, one could just as easily do any odd-numbered
power (except that typical calculators don't display enough digits!)

Round-Robin Tournament Schedules

This one came up when one of my students was trying to make a schedule
for a twelve-team pool competition. Essentially, she wanted every
team to be able to play every other team exactly once, each playing one
game per round, over an eleven-round competition.

For fewer teams, this is a simple exercise. For example, with
only two teams its trivial: Team 1 v Team 2. For four teams a suitable
schedule is:

Round 1
1 v 2
3 v 4

Round 2
1 v 3
2 v 4

Round 3
1 v 4
2 v 3

Six teams is a little trickier, but can be done by trial and error with pen
and paper in just a few minutes. As the number of teams in the competition
is increased, this becomes very difficult very quickly --- try doing it for just
10 teams!

Although the solution isn't unique (different schedules can be produced by
reordering rounds or relabelling teams), the total number of possible orderings
of these games is huge for even a small number of teams. If you had to do
it by trial and error, this would take an excessively long time.

Fortunately, there exist some simple algorithms for generating these
schedules quickly.